A machine gun fires a bullet of mass 35 gram with a speed of $600 \mathrm{~m} / \mathrm{s}$. The person holding the gum can extract a maximum force of 147 N on it. The number of bullets that can be fired from the gun per second is

The force exerted by the gun equals the rate of change of momentum of the bullets. For each bullet of mass $m = 0.035 \text{ kg}$ fired at speed $v = 600 \text{ m/s}$, the momentum per bullet is $mv$.

If $n$ bullets are fired per second, the rate of change of momentum is $n \cdot (m v)$.

Using Newton’s second law, $F = n m v$, with $F = 147 \text{ N}$:

$147 = n \cdot (0.035 \times 600)$

Since $0.035 \times 600 = 21$, we have:

$147 = 21n$

Solving for $n$:

$n = \frac{147}{21} = 7$

Final answer: $\boxed{7}$